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Group SL(2,C)

  1. Sep 2, 2010 #1
    Hi,

    I'm just reading about the group SL(2,C). In the book that I'm using(Jones, groups reps and physics), he defines a 2x2 matrix from a generic 4 vector [tex]v_{\mu}[/tex] and a vector [tex] \sigma_{\mu}:=(1,\vec{\sigma})[/tex], as [tex] V:=v_{\mu}\sigma^{\mu}[/tex]

    He nows wants to invert this equation to solve for [tex]v_{\mu}[/tex], and he suggests tracing with another vector of matrices defined as [tex] \tilde{\sigma_{\mu}}:=(1,-\vec{\sigma})[/tex], and he obtains [tex] v_{\mu}=\tfrac{1}{2}Tr(\tilde{\sigma_{\mu}}V)[/tex]

    I can't seem to get this, starting with [tex] V:=v_{\mu}\sigma^{\mu}[/tex] and then multiplying by [tex] \tilde{\sigma_{\nu}} [/tex], leads to [tex] \tilde{\sigma_{\nu}}V:=v_{\mu}\sigma^{\mu}\tilde{\sigma_{\nu}}[/tex]

    Now I'm not sure what indices I'm supposed to trace with?
     
  2. jcsd
  3. Sep 2, 2010 #2
    Re: Sl(2,c)

    Your V is a 2x2 matrix. If you want to recover [tex]v_{\nu}[/tex] (i.e. the nu'th component of the vector v), you should take [tex]\frac{1}{2} Tr(\tilde{\sigma}_{\nu}V)[/tex], where nu is not summed over. In other words, if you want the first component, take the 2x2 matrix [tex]\tilde{\sigma}_1[/tex], left-multiply (though inside a trace order doesn't matter) it with the 2x2 matrix V, and take half the trace of the resultant 2x2 matrix. Try a simple example if this isn't clear; pick (1,0,0,0) and try to recover v_0.
     
  4. Sep 2, 2010 #3

    Fredrik

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    Re: Sl(2,c)

    I'm going to write all indices as subscripts. [itex]\{I,\sigma_1,\sigma_2,\sigma_3\}[/itex], is a basis for the real vector space of complex 2×2 self-adjoint matrices. I'll call that space V. If we define an inner product by

    [tex]\langle A,B\rangle=\frac{1}{2}\operatorname{Tr}(A^\dagger B)[/tex]

    for all A,B in V, it's an orthonormal basis. So if we define [itex]\sigma_0=I[/itex], any [itex]x \in V[/itex] can be expressed as [itex]x=x_\mu \sigma_\mu[/itex], with [itex]x_\mu=\langle\sigma_\mu,x\rangle[/itex]. Note that all the [itex]x_\mu[/itex] are real. (This is implied by the facts that V is a real vector space and that [itex]\{\sigma_0,\sigma_1,\sigma_2,\sigma_3\}[/itex] is a basis of V).

    [tex]x_\mu=\langle\sigma_\mu,x\rangle=\frac{1}{2}\operatorname{Tr}(\sigma_\mu^\dagger x)=\frac{1}{2}(\sigma_\mu x)_{\nu\nu}[/tex]

    The map [itex]\mathbb R^4\ni (x_0,x_1,x_2,x_3)\mapsto x_\mu\sigma_\mu\in V[/itex] is an isomorphism. So V is isomorphic to [itex]\mathbb R^4[/itex].
     
    Last edited: Sep 2, 2010
  5. Sep 5, 2010 #4
    Re: Sl(2,c)

    Thanks for the help.

    I convinced myself of this in the end.

    [tex]
    \tilde{\sigma_{\mu}}V:=\tilde{\sigma_{\mu}}(v_{\beta}\sigma^{\beta})
    [/tex]
    Tracing (on the 2x2 matrix indices, not 4 vec indices, which is I think what was confusing me):
    [tex] Tr( \tilde{\sigma_{\mu}}V)=( \tilde{\sigma_{\mu}}V)_{ii}=(\tilde{\sigma_{\mu}})_{ij}(v_{\beta}\sigma^{\beta})_{ji}=v_{\beta} (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}[/tex]

    Now we have that, [tex] \tilde{\sigma_{\mu}}=(1,-\vec{\sigma}) [/tex] and [tex] \sigma^{\mu}=(1,-\vec{\sigma}) [/tex]. Therefore if [tex]\mu=\beta[/tex] : [tex] (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}=1_{ii}=2 [/tex] (since e.g. 1x1=1, [tex] \sigma_{x}\sigma_{x}=1 [/tex], [tex] \sigma_{y}\sigma_{y}=1 [/tex], [tex] \sigma_{z}\sigma_{z}=1 [/tex] etc, no summation)

    On the other hand if [tex]\mu\neq\beta[/tex], we end up with [tex] (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji} [/tex] equal to the trace on another Pauli matrix by virtue of the cyclic identity [tex] \sigma_{i}\sigma_{j}=i\epsilon_{ijk}\sigma_{k} [/tex], and the since the Pauli matrices are traceless, this trace of the product is zero.

    Combining these facts:

    [tex] Tr(\tilde{\sigma}_{\mu}V)=v_{\beta}\delta^{\beta}_{\mu}2 [/tex] which implies [tex] v_{\beta}=\tfrac{1}{2}Tr(\tilde{\sigma}_{\mu}V) [/tex]
     
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